# Kate, a bungee jumper, wants to jump off the edge of a bridge that spans a river below. Kate has a mass m, and the surface of the bridge is a height h above the water. The bungee cord, which has length L when unstretched, will first straighten and then stretch as Kate falls. Assume the following: The bungee cord behaves as an ideal spring once it begins to stretch, with spring constant k. Kate doesn't actually jump but simply steps off the edge of the bridge and falls straight downward. Kate's height is negligible compared to the length of the bungee cord. Hence, she can be treated as a point particle. Use g for the magnitude of the acceleration due to gravity.

Point motion will eventually stops due to action of g exactly perpendicular...

Explanation:

If ignoring the air resistance, the magnitude of gravitational acceleration is already strong enough to stops the acceleration. As we know that, the spring constant of a bungee spring cord will be F = -k/x, where x is the stretched length and k is the spring constant of bungee cord. If F = ma = w = mg, the g = -m  k/x. Now we can clearly see that the value of g remains constant due to the fluctuating length of the cord as the motion progresses back and forth in SHM say from x1 to x2 and x2 to x1.

## Related Questions

A simple pendulum takes 2.20 s to make one compete swing. If we now triple the length, how long will it take for one complete swing?

Time taken for 1 swing = 3.81 second

Explanation:

Given:

Time taken for 1 swing = 2.20 Sec

Find:

Time taken for 1 swing , when triple the length(T2)

Computation:

Time taken for 1 swing = 2π[√l/g]

2.20 = 2π[√l/g].......Eq1

Time taken for 1 swing , when triple the length (3L)

Time taken for 1 swing = 2π[√3l/g].......Eq2

Squaring and dividing the eq(1) by (2)

4.84 / T2² = 1 / 3

T2 = 3.81 second

Time taken for 1 swing = 3.81 second

A weather balloon is designed to expand to a maximum radius of 24 m at its working altitude, where the air pressure is 0.030 atm and the temperature is 200 K. If the balloon is filled at atmospheric pressure and 349 K, what is its radius at liftoff

Explanation:

At the working altitude;

air pressure = 0.030 atm

air temperature = 200 K

At liftoff;

temperature = 349 K

pressure = 1 atm

First, we assume balloon is spherical in nature,

and that the working gas obeys the gas laws.

from the radius, we can find the volume of the balloon at working atmosphere.

Volume of a sphere =

volume of balloon = x 3.142 x = 57913.34 m^3

using the gas equation,

=

The subscript 1 indicates the properties of the gas at working altitude, and the subscript 2 indicates properties of the gas at liftoff.

imputing values, we have

=

0.03 x 57913.34 x 349 = 200V2

V2 = 606352.67/200 = 3031.76 m^3  this is the volume occupied by the gas in the balloon at liftoff.

from the formula volume of a sphere,

V =   =   x 3.142 x = 3031.76

4.19  = 3031.76

= 3031.76/4.19

radius r of the balloon on liftoff = = 8.98 m

Label the longitudinal wave

???

Explanation:

The boom is supported by the winch cable that has a diameter of 0.5 in. and allowable normal stress of σallow=21 ksi. A boom rises from pin A to B, at angle phi from horizontal. A cable goes up from the load to B and then left to a pulley. Determine the greatest weight of the crate that can be supported without causing the cable to fail if ϕ=30∘. Neglect the size of the winch.

Explanation:

Let us assume that forces acting at point B are as follows.

= 0 ...... (1)

= 0

= 0 .......... (2)

Hence, formula for allowable normal stress of cable is as follows.

T =

= 3925 kip

From equation (1),    = -3925

= -3925

= 12877.29 kip

From equation (2),    -12877.29 (Cos 60) + W = 0

= 0

W = 6438.64 kip

Thus, we can conclude that greatest weight of the crate is 6438.64 kip.

To determine the greatest weight of the crate that can be supported without causing the cable to fail, calculate the normal stress on the cable using σ = F/A, where σ is the normal stress, F is the force on the cable, and A is the cross-sectional area of the cable. Then, compare the calculated normal stress to the allowable normal stress. Consider the angle phi in this calculation by using the equation F = W / sin(ϕ), where F is the force on the cable, W is the weight of the crate, and ϕ is the angle with respect to the horizontal.

To determine the greatest weight of the crate that can be supported without causing the cable to fail, we need to calculate the normal stress on the cable and compare it to the allowable normal stress. The normal stress can be calculated using the formula σ = F/A, where σ is the normal stress, F is the force on the cable, and A is the cross-sectional area of the cable. In this case, the force on the cable is equal to the weight of the crate, and the cross-sectional area of the cable can be calculated using the formula A = π*(d/2)^2, where d is the diameter of the cable.

Given that the diameter of the cable is 0.5 in and the allowable normal stress is 21 ksi, we can substitute these values into the equations and solve for the force on the cable:

Calculate the cross-sectional area of the cable: A = π*(0.5/2)^2 = π*(0.25)^2 = 0.1963 in^2

Plug the cross-sectional area and the allowable normal stress into the formula for normal stress: σallow = F/A → 21 ksi = F/0.1963 in^2

Solve for the force on the cable: F = 21 ksi * 0.1963 in^2 = 4.1183 ksi*in^2

Therefore, the greatest weight of the crate that can be supported without causing the cable to fail is equal to the force on the cable, which is 4.1183 ksi*in^2. However, it's important to note that we also need to consider the angle phi (ϕ) in this calculation. Since the cable goes up from the load to point B and then left to a pulley, the weight of the crate will create a vertical component and a horizontal component. To determine the weight of the crate that corresponds to the calculated force on the cable, we need to consider the trigonometric relationship between the force and the weight at an angle. In this case, the angle is 30 degrees, so we can use the equation F = W / sin(ϕ), where F is the force on the cable, W is the weight of the crate, and ϕ is the angle with respect to the horizontal.

Given that ϕ = 30 degrees and F = 4.1183 ksi*in^2, we can substitute these values into the equation and solve for the weight of the crate:

Plug the values into the equation: 4.1183 ksi*in^2 = W / sin(30)

Solve for the weight of the crate: W = 4.1183 ksi*in^2 * sin(30)

Therefore, the greatest weight of the crate that can be supported without causing the cable to fail and at an angle of 30 degrees is equal to the force on the cable, which is 4.1183 ksi*in^2, multiplied by the sine of 30 degrees.

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What is the magnitude and direction of the electric field atradiaConsider a coaxial conducting cable consisting of a conductingrod of radius R1 inside of a thin-walled conducting shell of radius 2(both are infinite length). Suppose the inner rod hasradiusR1= 1.3 mm and outer shell has radiusR2= 10R1Ifthe net charge density on the center rod isq1= 3.4×10−12C/mand the outer shell isq2=−2q1,a.)What is the magnitude and direction of the electric field atradial distancer= 5R1from the center rod

E = 9.4 10⁶ N / C,     The field goes from the inner cylinder to the outside

Explanation:

The best way to work this problem is with Gauss's law

Ф = E. dA = qint / ε₀

We must define a Gaussian surface, which takes advantage of the symmetry of the problem. We select a cylinder with the faces perpendicular to the coaxial.

The flow on the faces is zero, since the field goes in the radial direction of the cylinders.

The area of ​​the cylinder is the length of the circle along the length of the cable

dA = 2π dr L

A = 2π r L

They indicate that the distance at which we must calculate the field is

r = 5 R₁

r = 5 1.3

r = 6.5 mm

The radius of the outer shell is

r₂ = 10 R₁

r₂ = 10 1.3

r₂ = 13 mm

r₂ > r

When comparing these two values ​​we see that the field must be calculated between the two housings.

Gauss's law states that the charge is on the outside of the Gaussian surface does not contribute to the field, the charged on the inside of the surface is

λ = q / L

Qint = λ L

Let's replace

E 2π r L = λ L /ε₀

E = 1 / 2piε₀  λ / r

Let's calculate

E = 1 / 2pi 8.85 10⁻¹²  3.4 10-12 / 6.5 10-3

E = 9.4 10⁶ N / C

The field goes from the inner cylinder to the outside

You hit a hockey puck and it slides across the ice at nearly a constant speed.Is a force keeping it in motion?Explain.

At constant speed and varying position of the hockey puck, implies a change in the velocity of the hockey puck and net force is acting on it to keep it in motion.

According to Newton's second law of motion, the force applied to a an object is directly proportional to the product of mass and acceleration of the object.

F = ma

Acceleration is the change in the velocity of an object per change in time of motion.

• At constant velocity, the acceleration of an object iszero.
• When acceleration of an object is zero, the force on the object is zero.
• A constant speed (magnitude only) and change in the direction of the object, implies a change in velocity of the object.
• at changing velocity, the acceleration on an object is positive, and hence net force acts on the object.

Thus, we can conclude that at constant speed and varying position of the hockey puck, implies a change in the velocity of the hockey puck and net force is acting on it to keep it in motion.